For what values of $x$ does the series converge: $sum limits_{n=1}^{infty} frac{x^n}{n^n}$?
$begingroup$
For what values of $x$ do the following series converge or diverge
$$sum limits_{n=1}^{infty} frac{x^n}{n^n}$$
I tried to solve this using the ratio test where the series converge when
$$lim limits_{n to infty} frac{x^{n+1}n^n}{(n+1)^{n+1}x^n} <1$$
$$lim limits_{n to infty} frac{xn^n}{(n+1)^{n+1}} <1$$
but then I am not sure what to do next.
Please give me some ideas or hints on how to solve this question, thanks to anybody who helps.
real-analysis sequences-and-series
$endgroup$
add a comment |
$begingroup$
For what values of $x$ do the following series converge or diverge
$$sum limits_{n=1}^{infty} frac{x^n}{n^n}$$
I tried to solve this using the ratio test where the series converge when
$$lim limits_{n to infty} frac{x^{n+1}n^n}{(n+1)^{n+1}x^n} <1$$
$$lim limits_{n to infty} frac{xn^n}{(n+1)^{n+1}} <1$$
but then I am not sure what to do next.
Please give me some ideas or hints on how to solve this question, thanks to anybody who helps.
real-analysis sequences-and-series
$endgroup$
$begingroup$
Is $x$ assumed to be real or complex?
$endgroup$
– Luis Mendo
Nov 30 '14 at 3:44
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@Luis Mendo $x$ is assumed to be real.
$endgroup$
– Lucy
Nov 30 '14 at 3:47
$begingroup$
@Lucy, won't the root test easier to apply here?
$endgroup$
– abel
Nov 30 '14 at 4:18
add a comment |
$begingroup$
For what values of $x$ do the following series converge or diverge
$$sum limits_{n=1}^{infty} frac{x^n}{n^n}$$
I tried to solve this using the ratio test where the series converge when
$$lim limits_{n to infty} frac{x^{n+1}n^n}{(n+1)^{n+1}x^n} <1$$
$$lim limits_{n to infty} frac{xn^n}{(n+1)^{n+1}} <1$$
but then I am not sure what to do next.
Please give me some ideas or hints on how to solve this question, thanks to anybody who helps.
real-analysis sequences-and-series
$endgroup$
For what values of $x$ do the following series converge or diverge
$$sum limits_{n=1}^{infty} frac{x^n}{n^n}$$
I tried to solve this using the ratio test where the series converge when
$$lim limits_{n to infty} frac{x^{n+1}n^n}{(n+1)^{n+1}x^n} <1$$
$$lim limits_{n to infty} frac{xn^n}{(n+1)^{n+1}} <1$$
but then I am not sure what to do next.
Please give me some ideas or hints on how to solve this question, thanks to anybody who helps.
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Dec 8 '18 at 9:31
Martin Sleziak
44.7k9117272
44.7k9117272
asked Nov 30 '14 at 3:43
LucyLucy
495
495
$begingroup$
Is $x$ assumed to be real or complex?
$endgroup$
– Luis Mendo
Nov 30 '14 at 3:44
$begingroup$
@Luis Mendo $x$ is assumed to be real.
$endgroup$
– Lucy
Nov 30 '14 at 3:47
$begingroup$
@Lucy, won't the root test easier to apply here?
$endgroup$
– abel
Nov 30 '14 at 4:18
add a comment |
$begingroup$
Is $x$ assumed to be real or complex?
$endgroup$
– Luis Mendo
Nov 30 '14 at 3:44
$begingroup$
@Luis Mendo $x$ is assumed to be real.
$endgroup$
– Lucy
Nov 30 '14 at 3:47
$begingroup$
@Lucy, won't the root test easier to apply here?
$endgroup$
– abel
Nov 30 '14 at 4:18
$begingroup$
Is $x$ assumed to be real or complex?
$endgroup$
– Luis Mendo
Nov 30 '14 at 3:44
$begingroup$
Is $x$ assumed to be real or complex?
$endgroup$
– Luis Mendo
Nov 30 '14 at 3:44
$begingroup$
@Luis Mendo $x$ is assumed to be real.
$endgroup$
– Lucy
Nov 30 '14 at 3:47
$begingroup$
@Luis Mendo $x$ is assumed to be real.
$endgroup$
– Lucy
Nov 30 '14 at 3:47
$begingroup$
@Lucy, won't the root test easier to apply here?
$endgroup$
– abel
Nov 30 '14 at 4:18
$begingroup$
@Lucy, won't the root test easier to apply here?
$endgroup$
– abel
Nov 30 '14 at 4:18
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Staring from the last step you have shown we have
$$lim limits_{n to infty} frac{xn^n}{(n+1)^{n+1}} <1.$$
But
$$lim limits_{n to infty} frac{xn^n}{(n+1)^{n+1}} =lim_{n to infty} frac{x}{left(1+frac{1}{n}right)^n(n+1)}.$$
As $n to infty$, the limit $lim_{n to infty}left(1+frac{1}{n}right)^n=e.$ Thus the limit
$$lim_{n to infty} frac{x}{left(1+frac{1}{n}right)^n(n+1)}=0 quad forall x.$$
$endgroup$
$begingroup$
Yet the OP arrived to the first expression applying the ratio test, so the above proves only the series converges for all non-negative $;x;$ , whereas in fact it converges for any $;x;$ .
$endgroup$
– Timbuc
Nov 30 '14 at 4:02
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@Timbuc I'm not sure how you are restricting the ratio test to positive $x$ only. Because it should be based on $limleft|frac{a_n}{a_{n+1}}right|$?
$endgroup$
– Anurag A
Dec 8 '14 at 5:11
$begingroup$
I'm restricting nothing. I just made a point about what the OP did
$endgroup$
– Timbuc
Dec 8 '14 at 5:15
add a comment |
$begingroup$
Use test comparison against geometric series, by noting that for any $n>3|x|$ it holds
$$frac{|x|^n}{n^n}leq left(frac13right)^n$$
$endgroup$
add a comment |
$begingroup$
Hints
Upper-bound by a geometric series with ratio between 0 and 1 ... (hover mouse to see more)
That series will be convergent, and then yours will be assured to converge too because of the direct comparison test.
... (and then some more) ...
To obtain that convergent geometric series, you may need to remove some initial terms.
Detailed answer
I assume you ask about pointwise convergence, that is, convergence of the numeric series corresponding to a fixed value of $x$. I also assume that $x$ is complex in general.
Given $x$, let $n_0$ be an integer greater than $|x|$, and let $a = |x|/n_0$. Then $0<a<1$, and $|x|/n leq a$ for all $n geq n_0$. So, removing the $n_0-1$ initial terms (which doesn't affect convergence),
$$
left| sum_{n=n_0}^infty frac{x^n}{n^n} right|
leq sum_{n=n_0}^infty frac{|x|^n}{n^n}
leq sum_{n=n_0}^infty a^n.
$$
The latter is a geometric series, which is absolutely convergent because $0 < a <1$. So, applying the direct comparison test it stems that for any complex $x$, the original series is absolutely convergent.
$endgroup$
add a comment |
$begingroup$
Hint: $$dfrac{n^n}{(n+1)^{n+1}} = dfrac{1}{n+1} left( dfrac{n}{n+1}right)^n < ldots$$
$endgroup$
add a comment |
$begingroup$
For the ratio test you have to use absolute value as you're not said what $;x;$ , and this also works in the complex case. But in this case the n-th root test is better, I believe::
$$sqrt[n]{frac{|x|^n}{n^n}}=frac{|x|}nxrightarrow[ntoinfty]{}0$$
and the above is true for all $;xinBbb C;$ , so the series converges for all the complex.
$endgroup$
add a comment |
$begingroup$
Note that
$$
n^n ge n! implies frac{x^n}{n^n} le frac{x^n}{n!}
$$
and use the comparison test against the series formulation of $e^x$
Note the above only works for positive $x$, however consider showing that the series is absolutely convergent for any $x$ using the above inequalities and thus it is convergent for any $x$.
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Staring from the last step you have shown we have
$$lim limits_{n to infty} frac{xn^n}{(n+1)^{n+1}} <1.$$
But
$$lim limits_{n to infty} frac{xn^n}{(n+1)^{n+1}} =lim_{n to infty} frac{x}{left(1+frac{1}{n}right)^n(n+1)}.$$
As $n to infty$, the limit $lim_{n to infty}left(1+frac{1}{n}right)^n=e.$ Thus the limit
$$lim_{n to infty} frac{x}{left(1+frac{1}{n}right)^n(n+1)}=0 quad forall x.$$
$endgroup$
$begingroup$
Yet the OP arrived to the first expression applying the ratio test, so the above proves only the series converges for all non-negative $;x;$ , whereas in fact it converges for any $;x;$ .
$endgroup$
– Timbuc
Nov 30 '14 at 4:02
$begingroup$
@Timbuc I'm not sure how you are restricting the ratio test to positive $x$ only. Because it should be based on $limleft|frac{a_n}{a_{n+1}}right|$?
$endgroup$
– Anurag A
Dec 8 '14 at 5:11
$begingroup$
I'm restricting nothing. I just made a point about what the OP did
$endgroup$
– Timbuc
Dec 8 '14 at 5:15
add a comment |
$begingroup$
Staring from the last step you have shown we have
$$lim limits_{n to infty} frac{xn^n}{(n+1)^{n+1}} <1.$$
But
$$lim limits_{n to infty} frac{xn^n}{(n+1)^{n+1}} =lim_{n to infty} frac{x}{left(1+frac{1}{n}right)^n(n+1)}.$$
As $n to infty$, the limit $lim_{n to infty}left(1+frac{1}{n}right)^n=e.$ Thus the limit
$$lim_{n to infty} frac{x}{left(1+frac{1}{n}right)^n(n+1)}=0 quad forall x.$$
$endgroup$
$begingroup$
Yet the OP arrived to the first expression applying the ratio test, so the above proves only the series converges for all non-negative $;x;$ , whereas in fact it converges for any $;x;$ .
$endgroup$
– Timbuc
Nov 30 '14 at 4:02
$begingroup$
@Timbuc I'm not sure how you are restricting the ratio test to positive $x$ only. Because it should be based on $limleft|frac{a_n}{a_{n+1}}right|$?
$endgroup$
– Anurag A
Dec 8 '14 at 5:11
$begingroup$
I'm restricting nothing. I just made a point about what the OP did
$endgroup$
– Timbuc
Dec 8 '14 at 5:15
add a comment |
$begingroup$
Staring from the last step you have shown we have
$$lim limits_{n to infty} frac{xn^n}{(n+1)^{n+1}} <1.$$
But
$$lim limits_{n to infty} frac{xn^n}{(n+1)^{n+1}} =lim_{n to infty} frac{x}{left(1+frac{1}{n}right)^n(n+1)}.$$
As $n to infty$, the limit $lim_{n to infty}left(1+frac{1}{n}right)^n=e.$ Thus the limit
$$lim_{n to infty} frac{x}{left(1+frac{1}{n}right)^n(n+1)}=0 quad forall x.$$
$endgroup$
Staring from the last step you have shown we have
$$lim limits_{n to infty} frac{xn^n}{(n+1)^{n+1}} <1.$$
But
$$lim limits_{n to infty} frac{xn^n}{(n+1)^{n+1}} =lim_{n to infty} frac{x}{left(1+frac{1}{n}right)^n(n+1)}.$$
As $n to infty$, the limit $lim_{n to infty}left(1+frac{1}{n}right)^n=e.$ Thus the limit
$$lim_{n to infty} frac{x}{left(1+frac{1}{n}right)^n(n+1)}=0 quad forall x.$$
answered Nov 30 '14 at 3:49
Anurag AAnurag A
25.8k12249
25.8k12249
$begingroup$
Yet the OP arrived to the first expression applying the ratio test, so the above proves only the series converges for all non-negative $;x;$ , whereas in fact it converges for any $;x;$ .
$endgroup$
– Timbuc
Nov 30 '14 at 4:02
$begingroup$
@Timbuc I'm not sure how you are restricting the ratio test to positive $x$ only. Because it should be based on $limleft|frac{a_n}{a_{n+1}}right|$?
$endgroup$
– Anurag A
Dec 8 '14 at 5:11
$begingroup$
I'm restricting nothing. I just made a point about what the OP did
$endgroup$
– Timbuc
Dec 8 '14 at 5:15
add a comment |
$begingroup$
Yet the OP arrived to the first expression applying the ratio test, so the above proves only the series converges for all non-negative $;x;$ , whereas in fact it converges for any $;x;$ .
$endgroup$
– Timbuc
Nov 30 '14 at 4:02
$begingroup$
@Timbuc I'm not sure how you are restricting the ratio test to positive $x$ only. Because it should be based on $limleft|frac{a_n}{a_{n+1}}right|$?
$endgroup$
– Anurag A
Dec 8 '14 at 5:11
$begingroup$
I'm restricting nothing. I just made a point about what the OP did
$endgroup$
– Timbuc
Dec 8 '14 at 5:15
$begingroup$
Yet the OP arrived to the first expression applying the ratio test, so the above proves only the series converges for all non-negative $;x;$ , whereas in fact it converges for any $;x;$ .
$endgroup$
– Timbuc
Nov 30 '14 at 4:02
$begingroup$
Yet the OP arrived to the first expression applying the ratio test, so the above proves only the series converges for all non-negative $;x;$ , whereas in fact it converges for any $;x;$ .
$endgroup$
– Timbuc
Nov 30 '14 at 4:02
$begingroup$
@Timbuc I'm not sure how you are restricting the ratio test to positive $x$ only. Because it should be based on $limleft|frac{a_n}{a_{n+1}}right|$?
$endgroup$
– Anurag A
Dec 8 '14 at 5:11
$begingroup$
@Timbuc I'm not sure how you are restricting the ratio test to positive $x$ only. Because it should be based on $limleft|frac{a_n}{a_{n+1}}right|$?
$endgroup$
– Anurag A
Dec 8 '14 at 5:11
$begingroup$
I'm restricting nothing. I just made a point about what the OP did
$endgroup$
– Timbuc
Dec 8 '14 at 5:15
$begingroup$
I'm restricting nothing. I just made a point about what the OP did
$endgroup$
– Timbuc
Dec 8 '14 at 5:15
add a comment |
$begingroup$
Use test comparison against geometric series, by noting that for any $n>3|x|$ it holds
$$frac{|x|^n}{n^n}leq left(frac13right)^n$$
$endgroup$
add a comment |
$begingroup$
Use test comparison against geometric series, by noting that for any $n>3|x|$ it holds
$$frac{|x|^n}{n^n}leq left(frac13right)^n$$
$endgroup$
add a comment |
$begingroup$
Use test comparison against geometric series, by noting that for any $n>3|x|$ it holds
$$frac{|x|^n}{n^n}leq left(frac13right)^n$$
$endgroup$
Use test comparison against geometric series, by noting that for any $n>3|x|$ it holds
$$frac{|x|^n}{n^n}leq left(frac13right)^n$$
answered Nov 30 '14 at 3:47
MillyMilly
2,618612
2,618612
add a comment |
add a comment |
$begingroup$
Hints
Upper-bound by a geometric series with ratio between 0 and 1 ... (hover mouse to see more)
That series will be convergent, and then yours will be assured to converge too because of the direct comparison test.
... (and then some more) ...
To obtain that convergent geometric series, you may need to remove some initial terms.
Detailed answer
I assume you ask about pointwise convergence, that is, convergence of the numeric series corresponding to a fixed value of $x$. I also assume that $x$ is complex in general.
Given $x$, let $n_0$ be an integer greater than $|x|$, and let $a = |x|/n_0$. Then $0<a<1$, and $|x|/n leq a$ for all $n geq n_0$. So, removing the $n_0-1$ initial terms (which doesn't affect convergence),
$$
left| sum_{n=n_0}^infty frac{x^n}{n^n} right|
leq sum_{n=n_0}^infty frac{|x|^n}{n^n}
leq sum_{n=n_0}^infty a^n.
$$
The latter is a geometric series, which is absolutely convergent because $0 < a <1$. So, applying the direct comparison test it stems that for any complex $x$, the original series is absolutely convergent.
$endgroup$
add a comment |
$begingroup$
Hints
Upper-bound by a geometric series with ratio between 0 and 1 ... (hover mouse to see more)
That series will be convergent, and then yours will be assured to converge too because of the direct comparison test.
... (and then some more) ...
To obtain that convergent geometric series, you may need to remove some initial terms.
Detailed answer
I assume you ask about pointwise convergence, that is, convergence of the numeric series corresponding to a fixed value of $x$. I also assume that $x$ is complex in general.
Given $x$, let $n_0$ be an integer greater than $|x|$, and let $a = |x|/n_0$. Then $0<a<1$, and $|x|/n leq a$ for all $n geq n_0$. So, removing the $n_0-1$ initial terms (which doesn't affect convergence),
$$
left| sum_{n=n_0}^infty frac{x^n}{n^n} right|
leq sum_{n=n_0}^infty frac{|x|^n}{n^n}
leq sum_{n=n_0}^infty a^n.
$$
The latter is a geometric series, which is absolutely convergent because $0 < a <1$. So, applying the direct comparison test it stems that for any complex $x$, the original series is absolutely convergent.
$endgroup$
add a comment |
$begingroup$
Hints
Upper-bound by a geometric series with ratio between 0 and 1 ... (hover mouse to see more)
That series will be convergent, and then yours will be assured to converge too because of the direct comparison test.
... (and then some more) ...
To obtain that convergent geometric series, you may need to remove some initial terms.
Detailed answer
I assume you ask about pointwise convergence, that is, convergence of the numeric series corresponding to a fixed value of $x$. I also assume that $x$ is complex in general.
Given $x$, let $n_0$ be an integer greater than $|x|$, and let $a = |x|/n_0$. Then $0<a<1$, and $|x|/n leq a$ for all $n geq n_0$. So, removing the $n_0-1$ initial terms (which doesn't affect convergence),
$$
left| sum_{n=n_0}^infty frac{x^n}{n^n} right|
leq sum_{n=n_0}^infty frac{|x|^n}{n^n}
leq sum_{n=n_0}^infty a^n.
$$
The latter is a geometric series, which is absolutely convergent because $0 < a <1$. So, applying the direct comparison test it stems that for any complex $x$, the original series is absolutely convergent.
$endgroup$
Hints
Upper-bound by a geometric series with ratio between 0 and 1 ... (hover mouse to see more)
That series will be convergent, and then yours will be assured to converge too because of the direct comparison test.
... (and then some more) ...
To obtain that convergent geometric series, you may need to remove some initial terms.
Detailed answer
I assume you ask about pointwise convergence, that is, convergence of the numeric series corresponding to a fixed value of $x$. I also assume that $x$ is complex in general.
Given $x$, let $n_0$ be an integer greater than $|x|$, and let $a = |x|/n_0$. Then $0<a<1$, and $|x|/n leq a$ for all $n geq n_0$. So, removing the $n_0-1$ initial terms (which doesn't affect convergence),
$$
left| sum_{n=n_0}^infty frac{x^n}{n^n} right|
leq sum_{n=n_0}^infty frac{|x|^n}{n^n}
leq sum_{n=n_0}^infty a^n.
$$
The latter is a geometric series, which is absolutely convergent because $0 < a <1$. So, applying the direct comparison test it stems that for any complex $x$, the original series is absolutely convergent.
edited Nov 30 '14 at 4:17
answered Nov 30 '14 at 3:48
Luis MendoLuis Mendo
1,3321821
1,3321821
add a comment |
add a comment |
$begingroup$
Hint: $$dfrac{n^n}{(n+1)^{n+1}} = dfrac{1}{n+1} left( dfrac{n}{n+1}right)^n < ldots$$
$endgroup$
add a comment |
$begingroup$
Hint: $$dfrac{n^n}{(n+1)^{n+1}} = dfrac{1}{n+1} left( dfrac{n}{n+1}right)^n < ldots$$
$endgroup$
add a comment |
$begingroup$
Hint: $$dfrac{n^n}{(n+1)^{n+1}} = dfrac{1}{n+1} left( dfrac{n}{n+1}right)^n < ldots$$
$endgroup$
Hint: $$dfrac{n^n}{(n+1)^{n+1}} = dfrac{1}{n+1} left( dfrac{n}{n+1}right)^n < ldots$$
answered Nov 30 '14 at 3:47
Robert IsraelRobert Israel
320k23209459
320k23209459
add a comment |
add a comment |
$begingroup$
For the ratio test you have to use absolute value as you're not said what $;x;$ , and this also works in the complex case. But in this case the n-th root test is better, I believe::
$$sqrt[n]{frac{|x|^n}{n^n}}=frac{|x|}nxrightarrow[ntoinfty]{}0$$
and the above is true for all $;xinBbb C;$ , so the series converges for all the complex.
$endgroup$
add a comment |
$begingroup$
For the ratio test you have to use absolute value as you're not said what $;x;$ , and this also works in the complex case. But in this case the n-th root test is better, I believe::
$$sqrt[n]{frac{|x|^n}{n^n}}=frac{|x|}nxrightarrow[ntoinfty]{}0$$
and the above is true for all $;xinBbb C;$ , so the series converges for all the complex.
$endgroup$
add a comment |
$begingroup$
For the ratio test you have to use absolute value as you're not said what $;x;$ , and this also works in the complex case. But in this case the n-th root test is better, I believe::
$$sqrt[n]{frac{|x|^n}{n^n}}=frac{|x|}nxrightarrow[ntoinfty]{}0$$
and the above is true for all $;xinBbb C;$ , so the series converges for all the complex.
$endgroup$
For the ratio test you have to use absolute value as you're not said what $;x;$ , and this also works in the complex case. But in this case the n-th root test is better, I believe::
$$sqrt[n]{frac{|x|^n}{n^n}}=frac{|x|}nxrightarrow[ntoinfty]{}0$$
and the above is true for all $;xinBbb C;$ , so the series converges for all the complex.
answered Nov 30 '14 at 3:49
TimbucTimbuc
30.9k22145
30.9k22145
add a comment |
add a comment |
$begingroup$
Note that
$$
n^n ge n! implies frac{x^n}{n^n} le frac{x^n}{n!}
$$
and use the comparison test against the series formulation of $e^x$
Note the above only works for positive $x$, however consider showing that the series is absolutely convergent for any $x$ using the above inequalities and thus it is convergent for any $x$.
$endgroup$
add a comment |
$begingroup$
Note that
$$
n^n ge n! implies frac{x^n}{n^n} le frac{x^n}{n!}
$$
and use the comparison test against the series formulation of $e^x$
Note the above only works for positive $x$, however consider showing that the series is absolutely convergent for any $x$ using the above inequalities and thus it is convergent for any $x$.
$endgroup$
add a comment |
$begingroup$
Note that
$$
n^n ge n! implies frac{x^n}{n^n} le frac{x^n}{n!}
$$
and use the comparison test against the series formulation of $e^x$
Note the above only works for positive $x$, however consider showing that the series is absolutely convergent for any $x$ using the above inequalities and thus it is convergent for any $x$.
$endgroup$
Note that
$$
n^n ge n! implies frac{x^n}{n^n} le frac{x^n}{n!}
$$
and use the comparison test against the series formulation of $e^x$
Note the above only works for positive $x$, however consider showing that the series is absolutely convergent for any $x$ using the above inequalities and thus it is convergent for any $x$.
answered Nov 30 '14 at 3:51
DanZimmDanZimm
4,48811333
4,48811333
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$begingroup$
Is $x$ assumed to be real or complex?
$endgroup$
– Luis Mendo
Nov 30 '14 at 3:44
$begingroup$
@Luis Mendo $x$ is assumed to be real.
$endgroup$
– Lucy
Nov 30 '14 at 3:47
$begingroup$
@Lucy, won't the root test easier to apply here?
$endgroup$
– abel
Nov 30 '14 at 4:18